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A207645 Triangle where T(n,k) = Product_{j=1..k} floor(n/j - 1), as read by rows n>=0, columns k=0..[n/2].

%I A207645 #15 Jun 12 2012 12:08:29
%S A207645 1,1,1,1,1,2,1,3,3,1,4,4,1,5,10,10,1,6,12,12,1,7,21,21,21,1,8,24,48,
%T A207645 48,1,9,36,72,72,72,1,10,40,80,80,80,1,11,55,165,330,330,330,1,12,60,
%U A207645 180,360,360,360,1,13,78,234,468,468,468,468,1,14,84,336,672,1344,1344,1344
%N A207645 Triangle where T(n,k) = Product_{j=1..k} floor(n/j - 1), as read by rows n>=0, columns k=0..[n/2].
%C A207645 Compare the definition to that of Pascal's triangle:
%C A207645 binomial(n,k) = Product_{j=1..k} ((n+1)/j - 1).
%H A207645 Paul D. Hanna, <a href="/A207645/b207645.txt">Rows n = 0..70, flattened</a>
%F A207645 Row sums equal A207643.
%F A207645 Antidiagonal sums form A207644.
%F A207645 Right border of even-indexed rows equals A207646.
%F A207645 Right border of odd-indexed rows equals A207647.
%e A207645 Triangle begins with row n=0 as:
%e A207645 1;
%e A207645 1;
%e A207645 1,  1;
%e A207645 1,  2;
%e A207645 1,  3,   3;
%e A207645 1,  4,   4;
%e A207645 1,  5,  10,  10;
%e A207645 1,  6,  12,  12;
%e A207645 1,  7,  21,  21,   21;
%e A207645 1,  8,  24,  48,   48;
%e A207645 1,  9,  36,  72,   72,   72;
%e A207645 1, 10,  40,  80,   80,   80;
%e A207645 1, 11,  55, 165,  330,  330,  330;
%e A207645 1, 12,  60, 180,  360,  360,  360;
%e A207645 1, 13,  78, 234,  468,  468,  468,  468;
%e A207645 1, 14,  84, 336,  672, 1344, 1344, 1344;
%e A207645 1, 15, 105, 420, 1260, 2520, 2520, 2520, 2520;
%e A207645 1, 16, 112, 448, 1344, 2688, 2688, 2688, 2688; ...
%t A207645 t[n_, k_] := Product[Floor[n/j - 1], {j, 1, k}]; Flatten[Table[t[n, k], {n, 0, 15}, {k, 0, Floor[n/2]}]] (* _Jean-François Alcover_, Jun 12 2012 *)
%o A207645 (PARI) {T(n,k)=if(k==0,1,prod(j=1,k,floor(n/j-1)))}
%o A207645 for(n=0,12,for(k=0,n\2,print1(T(n,k),", "));print(""))
%Y A207645 Cf. A207643, A207644, A207646, A207647.
%K A207645 nonn,nice,tabf
%O A207645 0,6
%A A207645 _Paul D. Hanna_, Feb 20 2012